A smooth vector bundle $p:E rightarrow B$ is a submersion
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Prove that a smooth vector bundle $p:E rightarrow B$ is a submersion. With $B$ being a smooth $k$ dimensional manifold, and $E$ be a bundle of dimension $n$.
I would like to see how one prove this. Below is my attempt, I would be happy for some comments too.
My proof relies on Theorem 4.14, pg 83, of Lee's Manifold,
(Global rank theorem). Let $M$ and $N$ be smooth manifolds, and suppose $F:M rightarrow N$ be a smooth map of constant rank. If $F$ is surjective, then $F$ is a smooth submersion.
It suffices to show that $p$ has constant rank. But with a local trivialization of $E$, over local chart $U$ of $B$, we know that $p$ has the coordinate representation,
$$ (x^1,ldots, x^k, x^{k+1}, ldots, x^{k+n}) rightarrow (x^1, ldots, x^k) $$
The Jacobian matrix has rank $k$.
differential-geometry differential-topology vector-bundles
asked Jan 2 at 2:06
CL.CL.
2,2712925
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add a comment |
$begingroup$
Prove that a smooth vector bundle $p:E rightarrow B$ is a submersion. With $B$ being a smooth $k$ dimensional manifold, and $E$ be a bundle of dimension $n$.
I would like to see how one prove this. Below is my attempt, I would be happy for some comments too.
My proof relies on Theorem 4.14, pg 83, of Lee's Manifold,
(Global rank theorem). Let $M$ and $N$ be smooth manifolds, and suppose $F:M rightarrow N$ be a smooth map of constant rank. If $F$ is surjective, then $F$ is a smooth submersion.
It suffices to show that $p$ has constant rank. But with a local trivialization of $E$, over local chart $U$ of $B$, we know that $p$ has the coordinate representation,
$$ (x^1,ldots, x^k, x^{k+1}, ldots, x^{k+n}) rightarrow (x^1, ldots, x^k) $$
The Jacobian matrix has rank $k$.
differential-geometry differential-topology vector-bundles
asked Jan 2 at 2:06
CL.CL.
2,2712925
$endgroup$
add a comment |
$begingroup$
Prove that a smooth vector bundle $p:E rightarrow B$ is a submersion. With $B$ being a smooth $k$ dimensional manifold, and $E$ be a bundle of dimension $n$.
I would like to see how one prove this. Below is my attempt, I would be happy for some comments too.
My proof relies on Theorem 4.14, pg 83, of Lee's Manifold,
(Global rank theorem). Let $M$ and $N$ be smooth manifolds, and suppose $F:M rightarrow N$ be a smooth map of constant rank. If $F$ is surjective, then $F$ is a smooth submersion.
It suffices to show that $p$ has constant rank. But with a local trivialization of $E$, over local chart $U$ of $B$, we know that $p$ has the coordinate representation,
$$ (x^1,ldots, x^k, x^{k+1}, ldots, x^{k+n}) rightarrow (x^1, ldots, x^k) $$
The Jacobian matrix has rank $k$.
differential-geometry differential-topology vector-bundles
asked Jan 2 at 2:06
CL.CL.
2,2712925
$endgroup$
Prove that a smooth vector bundle $p:E rightarrow B$ is a submersion. With $B$ being a smooth $k$ dimensional manifold, and $E$ be a bundle of dimension $n$.
I would like to see how one prove this. Below is my attempt, I would be happy for some comments too.
My proof relies on Theorem 4.14, pg 83, of Lee's Manifold,
(Global rank theorem). Let $M$ and $N$ be smooth manifolds, and suppose $F:M rightarrow N$ be a smooth map of constant rank. If $F$ is surjective, then $F$ is a smooth submersion.
It suffices to show that $p$ has constant rank. But with a local trivialization of $E$, over local chart $U$ of $B$, we know that $p$ has the coordinate representation,
$$ (x^1,ldots, x^k, x^{k+1}, ldots, x^{k+n}) rightarrow (x^1, ldots, x^k) $$
The Jacobian matrix has rank $k$.
differential-geometry differential-topology vector-bundles
differential-geometry differential-topology vector-bundles
asked Jan 2 at 2:06
CL.CL.
2,2712925
asked Jan 2 at 2:06
CL.CL.
2,2712925
asked Jan 2 at 2:06
CL.CL.
2,2712925
asked Jan 2 at 2:06
CL.CL.
2,2712925
asked Jan 2 at 2:06
CL.CL.
2,2712925
2,2712925
add a comment |
add a comment |
1 Answer
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Your proof is fine but kind of silly. You have computed that the differential has rank $k=dim B$ at every point, which is the definition of what it means for $p$ to be a submersion. There's no need to go through the theorem you mentioned.
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
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$begingroup$
Oh yes, that is silly...
$endgroup$
– CL.
Jan 2 at 2:58
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
Your proof is fine but kind of silly. You have computed that the differential has rank $k=dim B$ at every point, which is the definition of what it means for $p$ to be a submersion. There's no need to go through the theorem you mentioned.
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
Oh yes, that is silly...
$endgroup$
– CL.
Jan 2 at 2:58
add a comment |
$begingroup$
Your proof is fine but kind of silly. You have computed that the differential has rank $k=dim B$ at every point, which is the definition of what it means for $p$ to be a submersion. There's no need to go through the theorem you mentioned.
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
Oh yes, that is silly...
$endgroup$
– CL.
Jan 2 at 2:58
add a comment |
$begingroup$
Your proof is fine but kind of silly. You have computed that the differential has rank $k=dim B$ at every point, which is the definition of what it means for $p$ to be a submersion. There's no need to go through the theorem you mentioned.
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
$endgroup$
Your proof is fine but kind of silly. You have computed that the differential has rank $k=dim B$ at every point, which is the definition of what it means for $p$ to be a submersion. There's no need to go through the theorem you mentioned.
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
answered Jan 2 at 2:57
Eric WofseyEric Wofsey
188k14216346
188k14216346
$begingroup$
Oh yes, that is silly...
$endgroup$
– CL.
Jan 2 at 2:58
add a comment |
$begingroup$
Oh yes, that is silly...
$endgroup$
– CL.
Jan 2 at 2:58
$begingroup$
Oh yes, that is silly...
$endgroup$
– CL.
Jan 2 at 2:58
$begingroup$
Oh yes, that is silly...
$endgroup$
– CL.
Jan 2 at 2:58
add a comment |
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